Boron’s chemistry is not typical of its group. is group 3A (13) shows the increasing metallic character from Al to Tl.
All Boron compounds are covalent whereas the other elements in group 3A (13) form mostly ionic compounds.
Except for Boron, the other elements of group 3A (13) show increasing metallic character from Al to Tl. But Boron is a metalloid.
Compared to the other elements in group 3A, boron has a lower reactivity in chemical terms (13)
The metalloid boron (B), as well as the metals aluminium (Al), gallium (Ga), indium (In), and thallium, are all part of group 3A (or IIIA) of the periodic table (Tl). In contrast to the other members of Group 3A, the element borax primarily forms covalent connections.
To learn more about group 3A (13) refer the link:
brainly.com/question/5489194
#SPJ4
Answer:
The electrode that removes ions from solution
Explanation:
Each electrochemical cell consists of an anode and a cathode. Oxidation occurs at the anode and reduction occurs at the cathode.
At the anode, ions move from the electrode into the solution while at the cathode ions move from the solution to the electrode.
At the cathode, metal ions accept electron(s) and become deposited on the electrode hence this electrode removes ions from solution. This is reduction.
Answer: C Snow
Explanation:
Because the temperature is low and it is below freezing temperature. Sorry if I am wrong.
Answer:
All igneous rocks the basis of the rock cycle are formed by plate tectonics. ... The heat from the mantle that fuels plate tectonics causes both igneous and sedimentary rocks to be turned into metamorphic rocks. The metamorphic rocks can be eroded into sedimentary rocks are remelted back into igneous.
Explanation:
The chemical equation is:
CH₄ + 2O₂ → CO₂ + 2H₂O
First, we calculate the moles of methane present using:
Moles = mass / molecular mass
Moles = 20 / 16
Moles = 1.25
Next, we may observe from the chemical equation that the molar ratio between methane and oxygen is 1 : 2
So the moles of oxygen required are 2 x 1.25
2.5 moles of oxygen required
Mass = moles * molecular mass
Mass = 2.5 * 32
Moles = 80
C. 80 grams O₂
